What is the area ABCD with vertices
A(-4,-6)
B(6, -6)
C(-1, 5)
D(9, 5)
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OpenStudy (anonymous):
okay. So you can either graph or use the distance formula and you have to find the distance between A and B and B and C then multiply. DO you know how to do that?
OpenStudy (anonymous):
whats the formula?
OpenStudy (anonymous):
@ab60093
OpenStudy (anonymous):
oh. Well, personally its easier to graph but the distance formula is
\[d=\sqrt{(x1-x2)^{2}+(y1-y2)^{2}}\]
OpenStudy (anonymous):
so i do that to each one? will it be the final answer?
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OpenStudy (anonymous):
You would only have to do it to AB and BC and no that wouldn't be the answer.so it would be set up like this for AB
\[d=\sqrt{(-4-6)^{2}+(-6-(-6))^{2}}\]
OpenStudy (anonymous):
and BC would be something like this
\[BC=\sqrt{(6-(-1))^{2}+(-6-5)^{2}}\]
OpenStudy (anonymous):
for AB its
\[AB=\sqrt{(10)^{2}+(0)^{2}}\]
OpenStudy (anonymous):
so the answer would be square root of 100?
OpenStudy (anonymous):
for AB
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OpenStudy (anonymous):
yes. For AB
OpenStudy (anonymous):
okay now for CD i do the same like AB?
OpenStudy (anonymous):
Not quite. You would do BC not CD. And i wrote the equation up top
OpenStudy (anonymous):
so it would be (7)^2+(11)^2?
OpenStudy (anonymous):
Correct!
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OpenStudy (anonymous):
so its \[\sqrt{170}\]
OpenStudy (anonymous):
i got 178
OpenStudy (anonymous):
its 7x7 which is 49 plus 11x11 which is 121. 121+49=170
OpenStudy (anonymous):
oh okay
OpenStudy (anonymous):
am i done yet?
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OpenStudy (anonymous):
not quite. You must now multiply the lengths of AB and BC. so
10 x square root of 170
OpenStudy (anonymous):
THAT would be the answer
OpenStudy (anonymous):
130.384048104?
OpenStudy (anonymous):
yeah. or just \[10\sqrt{170}\]
OpenStudy (anonymous):
another way to do this would have been the Pythagorean theorem but that would have been the same amount of work
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OpenStudy (anonymous):
omg thank you so much!
OpenStudy (anonymous):
No problem!
OpenStudy (anonymous):
i have one more problem though
OpenStudy (anonymous):
okay
OpenStudy (anonymous):
what is the area of DEF with vertices
D(-1, -5)
E(4, -5)
F(4, 7)
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OpenStudy (anonymous):
ooo. Triangle. fun
OpenStudy (anonymous):
ahaha
OpenStudy (anonymous):
okay. this one is going to be a little bit different to make our lives easier. We are going to use the graphing style. so when you do this you get a right triangle. do you know how to find the area of a right triangle?
OpenStudy (anonymous):
A=ab
2?
OpenStudy (anonymous):
a=ab/2?
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OpenStudy (anonymous):
yep. hold on like 2 seconds though
OpenStudy (anonymous):
okie
OpenStudy (anonymous):
okay. Sorry i was ordering food. ANYWAYS, uhhh the distance between E and F is 12 because...it just is. When you look at a graph the distance is 12, if you want we can do the distance formula to check later.
The distance between D and E is 5. You would now have to set up the formula
\[A=\frac{ d \times e }{ 2 }\]
OpenStudy (anonymous):
wait oops
OpenStudy (anonymous):
\[A=\frac{ DE \times EF }{ 2 }\]
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OpenStudy (anonymous):
so input values and solve
OpenStudy (anonymous):
okay hold on let me solve this lol
OpenStudy (anonymous):
so for D do i put -1 or -5?
OpenStudy (anonymous):
@ab60093
OpenStudy (anonymous):
oh, lol sorry
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OpenStudy (anonymous):
its alright
OpenStudy (anonymous):
it would be
\[A=\frac{ (5\times 12) }{ 2 }\]
OpenStudy (anonymous):
You wouldn't use the coordinates in the equation. You would use the distance between the coordinates